Given: Ry uses his AR-15 to shoot 1200 grams of Boomerite contained in a coffee creamer container. On top of the coffee creamer container is a 60 pound steel contraption for crushing charcoal briquettes and launching the dust into the air. Joe takes a video using his Windows Phone 7 phone and puts it up on YouTube*. In the video you can see the explosion occurred at 11.18 seconds into the video. The charcoal dispenser hits the ground at 14.48 seconds into the video. Afterward Ry measures the horizontal distance the charcoal dispenser traveled. It is 13 yards. Assume the acceleration of gravity on this planet at this location is 32.174 ft/sec2.
Problem: Ignoring air resistance and assuming the initial acceleration was for all practical purposes instantaneous answer the following questions:
- How high into the air did the charcoal dispenser go?
- At the instant after the detonation what were the horizontal and vertical velocity vectors of the charcoal dispenser?
- At the instant after the detonation what was the total velocity vector of the charcoal dispenser?
- What was the USPSA power factor of the charcoal dispenser at launch?
- If used at an USPSA match does the charcoal dispenser “make Major” for both pistol and rifle competition?
Be sure to use consistent units during the calculations and give the results in English units.
Solution:
- The total time in the air is 3.3 seconds. One half of the time is spent going up and the other half is spent going down. The equation of motion for an object dropped in a gravitational field is:
d = 1/2 a t2
Where d is the distance traveled in feet, a is the acceleration of the gravitational field, and t is the time in seconds.
The maximum height can be expressed as:
d = (32.174/2 ft/sec2) (3.3 sec/2)2
d = (16.087 ft/sec2)(1.65 sec)2
d = (16.087 ft/sec2)(2.7225 sec2)
d = 43.8 ft
- The equation of motion for an object traveling at a constant speed is:
d = v t
Where d is the distance traveled, v is the velocity, and t is the time.
This can be used to give us the initial horizontal velocity component.
Since the total time in the air was 3.3 seconds and the horizontal distance traveled was 13 yards the velocity can be solved for in the following equation:
13 yards = (v)(3.3 sec)
v = (13 yards)/(3.3 sec)
v = 3.94 yards/sec
or expressed in the more common feet per second:
v = (3 ft/yard)(3.94 yards/sec)
v = 11.8 ft/sec
The vertical component at launch is the same as the final vertical velocity at the moment of impact. The equation of velocity with respect to time is:
v = a t
Where v is the final velocity, a is acceleration, and t is the time.
Hence the initial vertical velocity is:
v = (32.174 ft/sec2)(3.3/2 sec)
v = (32.174 ft/sec2)(1.65 sec)
v = 53.1 ft/sec
- The total velocity is the square root of the sum of the squares of the horizontal and vertical velocities. Hence the total velocity at the instant after detonation was:
v = SQRT((11.8 ft/sec)2 + (53.1 ft/sec)2)
v = 54.4 ft/sec
- IPSC Power Factor is expressed by the following equations
PF = (m v)/1000
Where m is the mass of the bullet in grains and v is the velocity of the bullet in ft/sec.
There are 7000 grains in one pound. Hence the mass of the “bullet” is (7000)(60) or 420,000 grains.
Hence the IPSC Power Factor is:
PF = (420,000)(54.4)/1000
PF = 22,848
- The minimum USPSA power factor required to make major with a pistol is 165. For rifle it is 320. Since 22,848 is greater than both 165 and 320 the answer is “Yes”.
* The YouTube video:
Question: What was the secondary piece of black debris that flew back and wound up just behind and between the two individuals pictured?
A chunk of frozen sod.
We left a big crater in the field. Check out the last picture in this post. That is where the sod came from.
How high into the air did the charcoal dispenser go?
It went up the whole way.
At the instant after the detonation what were the horizontal and vertical velocity vectors of the charcoal dispenser?
Horizontal vector was not nearly as significant as vertical vector, which was (scientific terms) Up, Fast.
At the instant after the detonation what was the total velocity vector of the charcoal dispenser?
Again, up, fast.
What was the USPSA power factor of the charcoal dispenser at launch?
Six.
If used at an USPSA match does the charcoal dispenser “make Major” for both pistol and rifle competition?
Only if it hits the target–except to be exact, since there was no rifling used to impart spin on the projectile, then, no.
Show-off.
Projectile motion was my favorite part of physics. I think it was probably compounded by the fact we also had a potato gun in our dorm room. I still really want to build the control system I thought up for it. I want to swap it out for golf balls, for consistency, and have my own personal artillery piece. Alas, so many projects, so little time.
It’s impressive what that explosion did to the plate of AR500.
Joe, just imagine, if you hadn’t moved the car, it would be wearing that sod, and you could park it next to one of the high end cars at work. I love how the sod just sailed right between Ry and I.
Wolfwood,
The math is a very simple high school physics problem. Nothing to show off with here. I was just trying to provide an example so other people could do similar calculations when doing their own testing.
Barron,
Did you notice that you and Ry were both just inside the range of the debris and that I was just outside of it?
And Ry tried to make fun of me moving the vehicle…
🙂
I remember doing that exact calculation in high school. S=ut + 1/2at^2… v = u + at… v^2 = u^2 + 2as… oh, those were the days. Although we did everything in SI units, not feet.
By “that exact calculation”, I really mean that exact calculation – “A man fires a cannon in to the air. How long before it comes down? what height does it reach?” etc.
That chunk of sod was the equivalent of a chunk of pumpkin in speed, to mimic Ry, What’s the BC of a chunk of frozen sod. Solution step out of the way. My string of pictures actually stops from peak till just before impact because I was paying attention to debris.
You could have pulled the vehicle into the road and it would have been clean. I also think it was the size of that chunk that provided it with the distance it got.
I am amazed though by that chunk, I think that is the largest solid piece of dirt I’ve ever seen go flying with all the boomers I’ve shot and seen. I’ve seen stakes go flying, chunks of pumpkin go 100 yards but no large chunks of sod go airborne. Especially since the boomerite was just setting right on top of the snow, at least with the pumpkins the boomerite was inside.
That got me to wonder; what is the stored energy in just the spin of, say, a 52 grain .223 bullet at 300,000 R.P.M.? We calculate KE based on forward velocity alone, but in some cases there is a little more energy there to expend in the target if the bullet is on the ragged edge of letting go due to the centrifugal force. If I’m doing it correctly, I get a velocity, of tiny fragments letting go from the surface, of nearly 300 FPS. That would not be altogether insignificant, if, say, you had a stationary, spinning bullet let go right next to you at 300,000 RPM. I don’t know how to figure the total energy.